paper 17-17-2009. 305KB Jun 04 2011 12:03:07 AM
ACTA UNIVERSITATIS APULENSIS
No 17/2009
ABOUT BIHARMONIC PROBLEM VIA A SPECTRAL
APPROACH AND DECOMPOSITION TECHNIQUES
Lakhdar Benaissa, Noureddine Daili
Abstract. In this work, we describe a spectral method coupled with a
variational decomposition technique for solving a biharmonic equations. We
construct new spaces. Using one approximation by a spectral method we can
bring the resolution of Dirichlet problem for ∆2 to a finite number of Dirichlet
approach for -∆. Some new theoretic spectral approaches are given, numerical
solutions and illustrations are established to prove our theoretic study.
2000 Mathematics Subject Classification: 65N35, 65N30; 65N22.
Keywords and phrases: Biharmonic problem, spectral approach, approximation.
1. Introduction
Let Ω be a bounded domain of Rd (d = 1, 2, 3 in practice) of smooth boundary F r(Ω). We consider the following problem:
2
∆ u=f
in Ω,
u = g1
on F r(Ω) ;
(DP )
∂u
= g2 on F r(Ω) ,
∂η
where ∆2 = ∆(∆) is the biharmonic operator. The reduction of boundary
problems to equivalent problems is doing by several theoretic methods. We
are interested in this work to one direct method based on the Green formula.
This approach is used with success, by several authors, to resolve mathematical
modelling problems.
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L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
The solution of biharmonic problem (DP ) is studied by some authors and
by several methods, so by finite difference methods, finite element methods
and duality ...
J.W. McLaurin ([5], 1974) has given a decomposition technique for the
problem (DP ). He has proved that a solution of this problem by finite difference methods is equivalent to resolve a sequence of Dirichlet problems for the
operator ∆.
R. Glowinski and O. Pironneau ([4], 1977) have, only, established this decomposition method using the finite element methods correspondent to same
problem on a domain of R2 . They have proved that a solution of this problem
by this method is equivalent, also, to resolve a sequence of Dirichlet problems
for the operator ∆ and have given the error estimate:
ku − uh kH 1 (Ω) + k∆u + ϕh kL2 (Ω) =
(hk−1 ).
Our work consists to give a spectral approach to this problem and estimate
the error theoretically on new space. We illustrate our numerical approach by
numerical tests on some examples.
Indeed, if we resolve the biharmonic problem by spectral method based
on the direct variational formulation, then we can give as advantage of this
method the following :
i) we store one convergent approximation of the solution u in norm k.kH 1 (Ω)
in place of the norm k.kH 2 (Ω) ;
ii) We obtain a convergent approximation ϕN of −∆u.
However, we will think that, in spite of the simplicity of the space VN , the
practice problem is to compute (uN , ϕN ).
The basic idea in this case, consists to introduce a space MN ⊂ VN of
multipliers such that
VN = VN0 ⊕ MN
where
VN0 = {vN ∈ VN : vN = 0 on F r(Ω)}
In this paper, we prove that a resolution of Dirichlet problem for biharmonic
operator by Galerkin spectral method is equivalent to resolve a sequence of
Dirichlet problems for operator ∆ by the same method, and to resolve well
conditioned linear system. We prove some results of convergence which are
effecient and improve previously obtained results. We illustrate these results
by numerical trials.
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L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
2. A Continuous Problem
Let Ω be a bounded domain of Rd (d = 1, 2, 3 in practice) with smooth
boundary F r(Ω). Define the following problem:
2
∆ u=f
in Ω,
u = g1
on F r(Ω),
(DP )
∂u
= g2 on F r(Ω),
∂η
where g1 , g2 and f are given functions. This problem is equivalent to find
(u, w) such that
−∆w = f
in Ω,
−∆u = w on Ω,
(EP )
u = g1
on F r(Ω),
∂u
= g2 on F r(Ω).
∂η
3
1
We prove that if f ∈ L2 (Ω), g1 ∈ H 2 (F r(Ω)) and g2 ∈ H 2 (F r(Ω)) then (DP )
has one and only one solution in H 2 (Ω).
Theorem 1. [4] For all reals 0 ≤ ν ≤ µ we have
kukH µ (Ω) ≤ cN 2(µ−ν) kukH ν (Ω) ,
∀u ∈ VNd (Ω).
Theorem 2. [5] Define the space W 0 as follows:
W 0 = (v, ψ) ∈ H01 (Ω) × L2 (Ω), ∀µ ∈ H 1 (Ω), S((v, ψ), µ) = 0 ,
where
S((v, ψ), µ) =
Z
Ω
∇v∇µ dx −
Z
ψµ dx.
Ω
Then, a mapping : (v, ψ) ∈ W 0 → kψkL2 (Ω) is a norm on W 0 equivalent to the
norm:
(v, ψ) ∈ W 0 → (|v|2H 1 (Ω) + kψk2L2 (Ω) )1/2 .
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L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
Further
W 0 = (v, ψ) ∈ H02 (Ω) × L2 (Ω), − ∆v = ψ .
We have
1
Theorem 3. [6] Let λ ∈ H − 2 (F r(Ω)). The problem
2
∆ u = 0 in Ω,
−∆u = λ on F r(Ω),
u = 0 on F r(Ω),
(1)
has one and only one solution in H 2 (Ω)∩H01 (Ω), and the linear operator A
defined by
∂u
on F r(Ω)
Aλ = −
∂η
1
1
is an isomorphism from H − 2 (F r(Ω)) into H 2 (F r(Ω)). Also, the bilinear form
a(., .) defined by
a(λ, µ) =< Aλ, µ >
1
1
where < ., . > is the bilinear form of duality between H − 2 (F r(Ω)) and H 2 (F r(Ω)),
1
is continuous, symmetric and H − 2 (F r(Ω))-elliptic. Namely there exist an
α > 0 such that
a(µ, µ) ≥ α kµk2H − 21 (F r(Ω)) ,
1
∀µ ∈ H − 2 (F r(Ω)).
1
We will reduce the (DP ) problem to a variational equation in H − 2 (F r(Ω)).
Indeed, let w0 and u0 be solutions of
in Ω,
−∆w0 = f
(2)
w0 = 0
on F r(Ω);
−∆u0 = w0 in Ω,
(3)
u0 = g1 on F r(Ω).
We have then
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L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
Theorem 4. [4] Let u be a solution of (DP ). The trace λ of −∆u on
F r(Ω) is an unique solution of variational equation
< Aλ, µ >=<
∂u0
− g2 , µ > ,
∂η
1
1
∀µ ∈ H − 2 (F r(Ω)), λ ∈ H − 2 (F r(Ω)).
0
Remark 1. To determine ∂u
we resolve two Dirichlet problems (2), (3), and
∂η
we resolve more again two Dirichlet problems to obtain the couple (u, w).
3. Main Results
3.1. Approximation of (DP ) by Spectral Method
We consider the following finite dimension space:
VN = Span {L0 , L1 , ..., LN }
where Lk (x) are Legendre polynomials. Let
n
o
VN0 = vN ∈ VN : vN|F r(Ω) = 0
and let MN ⊂ VN be such that VN = VN0 ⊕ MN . We introduce the following
spaces :
and
(vN , ψN ) ∈ VN × VN , vN|F r(Ω) = g1 ,
WN =
;
R
R
R
∇vN ∇µN dx = Ω ψN µN dx + F r(Ω) g2 µN dσ, ∀µN ∈ VN
Ω
WN0
=
(vN , ψN ) ∈
VN0
× VN ,
Z
Ω
∇vN ∇µN dx −
Z
Ω
ψN µN dx = 0 , ∀µN ∈ VN
3.1.1. Choice of VN0 and Appropriate Basis for Galerkin Method
If we have a nonhomogenous Dirichlet problem associated to the operator
(∆) and posed on the space VN , we can define one transformation for which
this problem will be equivalent to one posed homogenous problem on VN0 .
The Galerkin approach consists plainly to replace the test functions space
by polynomials space.
The effectiveness of numerical method to resolve the linear system Au = F
wich has been given in the abstract form will be subordinate to :
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.
L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
i) the way from which space VN0 (Ω) approaches the space V ;
ii) steepness and the simpleness calculus of coefficients aij and Fj ;
iii) steepness to resolve a linear system Au = F.
To satisfy the 1st criterion i), we will consider the space VN0 (Ω) of
enough large dimension.
To satisfy the 2nd and 3rd critera ii) and iii), it will require obtain
one sufficiently deep matrix A such that the linear system Au = F
does not enough at cost (in time and required space machine), and
such that there are not coefficients aij to compute.
What is to be done in the choice of a basis of VN0 (Ω) such that the
linear system to resolve will be possible ?
To answer this question, we need the following lemma :
Lemma 1. [6] Put
ck =
√ 1
,
4k+6
′
φk (x) = ck (Lk (x) − Lk+2 (x))
′
ajk = (φk (x), φj (x))
,
bjk = (φk (x), φj (x)), k, j = 0, 1, ..., N − 2.
Then,
ajk =
1 if k = j,
0 if k 6= j,
2
2
+ 2j+5
),
if k = j,
ck cj ( 2j+1
=
−ck cj ,
if k = j + 2,
0
else,
bkj = bjk
and
VN0 (I) = Span {φ0 (x), φ1 (x), ..., φN −2 (x)} ,
where (., .) is an L2 -inner product.
3.1.2. Choice of MN
We select MN ⊂ VN such that VN = VN0 ⊕ MN , where
VN0 = Span {φ0 , φ1 , ..., φN −2 } .
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L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
Let φ−1 , φN −1 be two functions such that φ−1 , φN −1 ∈ VN and φ−1 ,φ0 ,
φ1 , ..., φN −2 , φN −1 are linearly independent.
Proposition 1. For d = 1, the space MN is given by
MN = Span {φ−1 (x), φN −1 (x)} .
For d = 2, the space MN is given by
φ−1 (x)φj (y), φN −1 (x)φj (y),
MN = Span
φi (x)φ−1 (y), φi (x)φN −1 (y),
−1≤j ≤N −1
0≤i≤N −2
.
For d = 3, the space MN is given by
φi (x)φj (y)φ−1 (z), φi (x)φj (y)φN −1 (z), −1 ≤ i, j ≤ N − 1;
φi (x)φ−1 (y)φk (z), φi (x)φN −1 (y)φk (z), −1 ≤ i ≤ N − 1,
MN = Span
0 ≤ k ≤ N − 2;
φ−1 (x)φj (y)φk (z), φN −1 (x)φj (y)φk (z), 0 ≤ j, k ≤ N − 2
Remark 2. The choice of functions φ−1 , φN −1 is not unique, then the
choice of space MN is not unique.
Now, we give some choices of φ−1 , φN −1 :
Remark 3. i) φ−1 = L0 , φN −1 = L1
ii) φ−1 = L0 , φN −1 = LN −1 − L1
iii) φ−1 = L1 − L0 , φN −1 = LN −1 .
Numerical tests are proofs to the good choice.
We approache then (DP ) by
Find (uN , ϕN ) ∈ WN such that
(DP )N
jN (uN , ϕN ) ≤ jN (vN , ψN ), ∀(vN , ψN ) ∈ WN ,
where
1
jN (vN , ψN ) =
2
Z
2
Ω
|ψN | dx −
This problem has one and only one solution.
163
Z
Ω
f vN dx.
.
L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
4. Convergence Results
Consider the following problem:
2
∆ u=g
u=0
(DP )0
∂u
=0
∂η
in Ω,
on F r(Ω),
on F r(Ω),
which is equivalent to the following optimization problem:
Find u ∈ H02 (Ω) such that
(DP )1
J0 (u) ≤ J0 (v), ∀v ∈ H02 (Ω),
where
1
J0 (v) =
2
Z
2
|∆v| dx −
Ω
Z
gvdx.
Ω
Now, we prove some results of convergence.
Lemma 2. Let u be a solution of problem (DP )0 . Then there exist a
constant c > 0 independent of N such that
|u − uN |H 1 (Ω) + k∆u + ϕN kL2 (Ω) ≤ c(
inf
0
(vN , ψN )∈WN
(|u − vN |H 1 (Ω) + k∆u + ψN kL2 (Ω) )
+ inf k∆u + µN kH 1 (Ω) ).
µN ∈VN
Proof. Let u be a solution of problem (DP )0 , then
Z
Z
Z
gvdx, ∀v ∈ D(Ω)
∆u∆vdx =
− ∇v∇ (∆u) dx =
Ω
Ω
Ω
but D(Ω) is dense in H01 (Ω), then
Z
Z
gvdx, ∀v ∈ H01 (Ω).
− ∇v∇(∆u)dx =
Ω
Ω
We have
S((v, ψ), µ) =
Z
Ω
∇v∇µdx −
164
Z
Ω
ψµdx,
L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
and
S((v, ψ), −∆u) = −
Z
Ω
∇v∇(∆u)dx +
Z
ψ∆udx =
Z
gvdx +
ψ∆udx.
Ω
Ω
Ω
Z
Let (vN , ψN ) ∈ WN0 and µN ∈ VN , then
R
S((uN − vN , ϕN − ψN ), ∆u + µN ) = Ω ∇(uN − vN )∇(∆u + µN )dx
−
R
Ω
(ϕN − ψN )(∆u + µN )dx
= S((uN − vN , ϕN − ψN ), ∆u) + S((uN − vN , ϕN − ψN ), µN )
R
=−
g(uN − vN )dx −
Ω
+
=−
R
Ω
+
Or
Z
Ω
R
Ω
∇uN ∇µN dx −
∇uN ∇µN dx =
Z
Ω
(ϕN − ψN )∆udx
∇(uN − vN )∇µN dx −
Ω
(ϕN − ψN )∆udx −
R
R
R
Ω
R
Ω
guN dx +
R
Ω
∇vN ∇µN dx −
ϕN µN dx and
Z
Ω
(ϕN − ψN )µN dx
gvN dx
R
Ω
ϕN µN dx +
ϕN ψN dx =
Ω
Ω
R
Z
R
Ω
ψN µN dx.
gvN dx,
Ω
then
Z
Ω
(ϕN − ψN )(∆u + ϕN )dx = −S((uN − vN , ϕN − ψN ), ∆u + µN )
By continuity of S ((., .) , .) , we have
R
(ϕN − ψN )(∆u + ϕN )dx ≤ c(|uN − vN | 1 k∆u + µN k 1
H (Ω)
H (Ω)
Ω
+ kϕN − ψN kL2 (Ω) k∆u + µN kH 1 (Ω) )
Using Theorem 2, it holds,
|v|H 1 (Ω) ≤ c(Ω) kψkL2 (Ω)
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L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
and
R
(ϕN − ψN )(∆u + ϕN )dx ≤ c(c(Ω) kϕN − ψN k 2 k∆u + µN k 1
L (Ω)
H (Ω)
Ω
+ kϕN − ψN kL2 (Ω) k∆u + µN kH 1 (Ω) )
≤ d(Ω) kϕN − ψN kL2 (Ω) k∆u + µN kH 1 (Ω) ,
where d(Ω) = max(cc(Ω), c). Thus
R
R
kϕN − ψN k2L2 (Ω) = Ω (ϕN − ψN )(∆u + ϕN )dx − Ω (ϕN − ψN )(∆u + ψN )dx
kϕN − ψN k2L2 (Ω) ≤ d(Ω) kϕN − ψN kL2 (Ω) k∆u + µN kH 1 (Ω)
+ kϕN − ψN kL2 (Ω) k∆u + ψN kL2 (Ω) .
It holds that
kϕN − ψN kL2 (Ω) ≤ d(Ω) k∆u + µN kH 1 (Ω) + k∆u + ψN kL2 (Ω)
(4)
and
|u − uN |H 1 (Ω) + k∆u + ϕN kL2 (Ω) ≤ |u − vN |H 1 (Ω) + |vN − uN |H 1 (Ω)
+ k∆u + ψN kL2 (Ω) + kϕN − ψN kL2 (Ω)
≤ |u − vN |H 1 (Ω) + k∆u + ψN kL2 (Ω) + (1 + c(Ω)) kϕN − ψN kL2 (Ω) .
Using inequality (4), we obtain
|u − uN |H 1 (Ω) + k∆u + ϕN kL2 (Ω) ≤ (|u − vN |H 1 (Ω) + k∆u + ψN kL2 (Ω)
+(1 + c(Ω))) · (d(Ω) k∆u + µN kH 1 (Ω) + k∆u + ψN kL2 (Ω) ),
hence
|u − uN |H 1 (Ω) + k∆u + ϕN kL2 (Ω) ≤ |u − vN |H 1 (Ω) + (2 + c(Ω)) k∆u + ψN kL2 (Ω)
+(1 + c(Ω))d(Ω) k∆u + µN kH 1 (Ω)
≤ c∗ (|u − vN |H 1 (Ω) + k∆u + ψN kL2 (Ω) + k∆u + µN kH 1 (Ω) ),
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L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
where
c∗ = max(2 + c(Ω), (1 + c(Ω))d(Ω)).
Therefore, it holds
|u − uN |H 1 (Ω) + k∆u + ϕN kL2 (Ω) ≤ c∗ (
inf
0
(vN ,ψN )∈WN
(|u − vN |H 1 (Ω) + k∆u + ψN kL2 (Ω) )
+ inf k∆u + µN kH 1 (Ω) ).
µN ∈VN
Lemma 3. Let u ∈ H 2 (Ω) be a solution of optimization problem (DP )0 .
If u ∈ H k+2 (Ω), k ≥ 2, then there exist a constant c > 0 independent of N
such that
|u − uN |H 1 (Ω) + k∆u + ϕN kL2 (Ω) ≤ cN 1−k (|u|H k+2 (Ω) + |∆u|H k (Ω) )
Proof. Let (vN , ψN ) ∈ WN0 and µN ∈ VN . Put
νN = µN + ψN , νN ∈ VN .
Then, we have
Z
Ω
Or
S((vN , ψN ), νN ) = 0,
Z
Z
∆uνN dx = − ∇u∇νN dx +
F r(Ω)
Ω
(5)
∂u
νN dσ
∂η
∂u
|F r(Ω) = 0, so one has
∂η
Z
Z
∆uνN dx = − ∇u∇νN dx.
Ω
Ω
By (5) and (6), we have
Z
Z
(∆u + ψN )νN dx =
∇(vN − u)∇νN dx
Ω
Ω
Z
(∆u + ψN )νN dx ≤ |u − vN | 1 |νN | 1 .
H (Ω)
H (Ω)
Ω
By Theorem 1, we obtain
|νN |H 1 (Ω) ≤ cN 2 kνN kL2 (Ω)
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(6)
L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
so
Z
(∆u + ψN )νN dx ≤ cN 2 |u − vN | 1 kνN k 2
H (Ω)
L (Ω)
Ω
And
kνN k2L2 (Ω) =
R
Ω
(µN − ∆u)νN dx +
R
Ω
(∆u + ψN )νN dx
≤ k∆u − µN kL2 (Ω) kνN kL2 (Ω) + cN 2 |u − vN |H 1 (Ω) kνN kL2 (Ω)
Hence
⇒ kνN kL2 (Ω) ≤ k∆u − µN kL2 (Ω) + cN 2 |u − vN |H 1 (Ω).
k∆u + ψN kL2 (Ω) = k∆u + νN − µN kL2 (Ω)
≤ k∆u − µN kL2 (Ω) + kνN kL2 (Ω) ≤ 2 k∆u − µN kL2 (Ω) + cN 2 |u − vN |H 1 (Ω) .
Then
|u − vN |H 1 (Ω) + k∆u + ψN kL2 (Ω) ≤ (1 + cN 2 ) |u − vN |H 1 (Ω) + 2 k∆u − µN kL2 (Ω) ,
∀ (vN , ψN ) ∈ WN0 , ∀µN ∈ VN .
Thus
inf
0
(vN ,ψN )∈WN
(|u − vN |H 1 (Ω) + k∆u + ψN kL2 (Ω) ) ≤ (1 + cN 2 ) inf 0 |u − vN |H 1 (Ω)
vN ∈VN
+2 inf k∆u − µN kL2 (Ω) .
µN ∈VN
Lemma 2 implies
|u − uN |H 1 (Ω) + k∆u + ϕN kL2 (Ω) ≤ c((1 + cN 2 ) inf 0 |u − vN |H 1 (Ω)
vN ∈VN
+ inf k∆u − µN kH 1 (Ω) ).
µN ∈VN
Further, we have
inf |u − vN |H 1 (Ω) ≤ cN −1−k |u|H k+2 (Ω)
vN ∈VN0
inf k∆u + µN kH 1 (Ω) ≤ cN 1−k |∆u|H k (Ω) ,
µN ∈VN
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L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
then, it holds
|u − uN |H 1 (Ω) + k∆u + ϕN kL2 (Ω) ≤ c((1 + cN 2 )cN −1−k |u|H k+2 (Ω) + cN 1−k |∆u|H k (Ω) )
≤ cN 1−k (|u|H k+2 (Ω) + |∆u|H k (Ω) ).
Theorem 5. Le u ∈ H 2 (Ω) be a solution of a problem (DP ), then for all
integer k ≥ 2, if u ∈ H k+2 (Ω) we have
ku − uN kH 1 (Ω) + k∆u + ϕN kL2 (Ω) ≤ cN 1−k kukH k+2 (Ω)
If u ∈ H k+1 (Ω), k ≥ 3, then
ku − uN kL2 (Ω) +
1
k∆u + ϕN kL2 (Ω) ≤ cN −1−k kukH k+1 (Ω)
N2
Proof. We will reduce the problem (P D)N to one variational problem in
MN . Indeed, let aN (., .) : MN × MN → R be a bilinear form defined for
λN ∈ MN by
Z
∇wN ∇vN dx = 0 , ∀vN ∈ VN0 , wN − λN ∈ VN0 ;
(7)
Ω
Z
Ω
∇uN ∇vN dx =
aN (λN , µN ) =
Z
Ω
Z
Ω
wN vN dx , ∀vN ∈ VN0 , uN ∈ VN0 ;
wN µN dx −
Z
Ω
∇uN ∇µN dx,
∀µN ∈ MN .
(8)
(9)
Lemma 4. A bilinear form aN (., .) is positive definite and symmetric on
MN × MN . Moreover
Z
w1N w2N dx, ∀ λ1N , λ2N ∈ MN ,
aN (λ1N , λ2N ) =
Ω
where w1N , w2N are solutions of (7) associated to λ1N , λ2N .
Proof. By definition one has
Z
Z
w1N λ2N dx − ∇u1N ∇λ2N dx, ∀µN ∈ MN ,
aN (λ1N , λ2N ) =
Ω
Ω
where w1N , w2N are solutions of (7) and (8) respectively.
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L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
Put λ2N = (λ2N − w2N ) + w2N , then
R
R
R
aN (λ1N , λ2N ) = Ω w1N w2N dx − Ω ∇u1N ∇w2N dx + Ω ∇u1N ∇(w2N − λ2N ) dx
−
R
Ω
w1N ∇(w2N − λ2N ) dx
One has u1N ∈ VN0 , using (7) one has
Z
Z
w1N ∇(w2N − λ2N ) dx.
∇u1N ∇(w2N − λ2N ) dx =
Ω
Ω
Then
aN (λ1N , λ2N ) =
Z
w1N w2N dx,
Ω
∀ λ1N , λ2N ∈ MN .
It is obvious that aN (., .) is symmetric and coercive.
Theorem 6. Let (uN , wN ) be a solution of (DP )N and λN the component of wN in MN . Then λN is an unique solution of the linear variational
problem:
R
R
R
aN (λN , µN ) = Ω ∇u0N ∇µN dx − Ω w0N µN dx − F r(Ω) g2N µN dσ,
(10)
∀µN ∈ MN , λN ∈ MN
where w0N is a solution of
Z
Z
f vN dx , ∀vN ∈ VN0 , w0N ∈ VN0
∇w0N ∇vN dx =
and u0N is a solution of
R
R
Ω ∇u0N ∇vN dx = Ω w0N vN dx , ∀vN ∈ VN0 ,
(11)
Ω
Ω
u0N = g1
on
F r(Ω).
Proof. We have
R
R
uN ∇µN dx
aN (λN , µN ) = Ω w¯N µN dx − Ω ∇¯
=
R
(wN − w0N )µN dx −
Ω
=
R
∇u0N ∇µN dx −
Ω
R
R
Ω
∇(uN − u0N )∇µN dx
R
R
w µ dx).
∇u
∇µ
dx
−
w
µ
dx
−
(
N
N
0N
N
Ω N N
Ω
Ω
170
(12)
L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
But (uN , wN ) ∈ WN , hence
Z
Z
Z
∇uN ∇µN dx − wN µN dx =
Ω
Ω
g2N µN dσ,
F r(Ω)
∀µN ∈ MN
Therefore
aN (λN , µN ) =
Z
Ω
Z
Z
∇u0N ∇µN dx − w0N µN dx−
Ω
g2N µN dσ,
F r(Ω)
∀µN ∈ MN .
Existence and uniqueness of a solution of (10) is a direct consequence from LaxMilgram Lemma. A system is symmetric because a form aN (., .) is symmetric.
Remak 4. The problem (10) is equivalent to a system from which associated matrix is positive definite and symmetric.
Now we prove a problem (10) is equivalent to a system from which associated matrix is positive definite.
5. Numerical Approach
5.1. Construction of Linear System Associated to the Variational
Problem (10)
Put
lN (µN ) =
Z
Ω
∇u0N ∇µN dx −
Z
Ω
w0N µN dx −
Z
g2 µN dσ,
F r(Ω)
∀µN ∈ MN
Then solve a problem (10) is equivalent to resolve the following system :
aN (λN , µN ) = lN (µN ),
∀µN ∈ MN , λN ∈ MN .
Define the set
{−1, N − 1}
(−1,
j),
(N
−
1,
j)
,
−1
≤
j
≤
N
−
1
,
(i, −1), (i, N − 1) ,
0≤i≤N −2
Id =
(i, j, −1), (i, j, N − 1) , −1 ≤ i, j ≤ N − 1,
(i, −1, k), (i, N − 1, k) , −1 ≤ i ≤ N − 1, 0 ≤ k ≤ N − 2
(−1, j, k), (N − 1, j, k), 0 ≤ j, k ≤ N − 2
171
(13)
if d = 1
if d = 2
, if d = 3,
L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
and
φid = φi1 φi2 · ... · φid
d = 1, 2, 3
where id = (i1 , i2 , ..., id ). So (13) is equivalent to
d
aN (λN , φid ) = lN (φid ) , i ∈ Id
P
λj d φj d .
λN =
(14)
j d ∈Id
Namely
X
λj d aN (φj d , φkd ) = lN (φkd ),
j d ∈Id
and
2,
4N,
p = Card(Id ) = dim(MN ) =
6N 2 + 2,
k d ∈ Id
(15)
if d = 1,
if d = 2,
if d = 3.
5.2. Computation of the Matrix AN
Denote BN the basis of MN and wj d N , uj d N solutions respectively of
R
wj d N ∈ VN , wj d N − φj d ∈ VN0 ;
∇wj d N ∇vN dx = 0, ∀vN ∈ VN0 ,
Ω
R
∇uj d N ∇vN dx =
Ω
further we have
R
Ω
aid j d = aN (φj d , φid ) =
wj d N vN dx, ∀vN ∈ VN0 ,
Z
Ω
wj d N φid dx −
Z
Ω
uj d N ∈ VN0 ,
∇uj d N ∇φid dx,
id , j d ∈ Id .
It results that a computation of a matrix AN requires the computation of 2p
Dirichlet problems for an operator ∆.
We have
Theorem 7. For all integer N ≥ 1, we have
α
1
kγ0 λN k2L2 (F r(Ω)) ≤ aN (λN , λN ) ≤ β kγ0 λN k2L2 (F r(Ω)) ,
N
where α and β are independents of N and λN .
172
∀λN ∈ MN ,
L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
Proof. Let λN ∈ MN , we have
aN (λN , λN ) =
Z
Ω
2
wN
dx,
where wN is a solution of (7). Let w˜N ∈ H 1 (Ω) be a solution of
∆w˜N = 0 in Ω,
w˜N = λN on F r (Ω) ,
then one has
p
aN (λN , λN ) = kwN kL2 (Ω)
≤ kwN − w˜N kL2 (Ω) + kw˜N kL2 (Ω) = kwN − w˜N kL2 (Ω) +
√
< Aγ0 λN , γ0 λN >.
Put |A| = kAkLc(L2 (F r(Ω)),L2 (F r(Ω))) , it holds that
|< Aλ, µ >| ≤ |A| kλkL2 (F r(Ω)) kµkL2 (F r(Ω)) ,
⇒
∀ λ, µ ∈ L2 (F r(Ω))
p
p
aN (λN , λN ) ≤ kwN − w˜N kL2 (Ω) + |A| kγ0 λN kL2 (F r(Ω)) .
and we have
3
kwN − w˜N kL2 (Ω) ≤ cN − 2 kw˜N kH 23 (Ω) ,
and
kw˜N kH 32 (Ω) ≤ c kλN kH 1 (F r(Ω))
kλN kH 1 (F r(Ω)) ≤ cN kλN kL2 (F r(Ω)) .
Namely
p
And
c
aN (λN , λN ) ≤ c kγ0 λN kL2 (F r(Ω)) .
1
kγ0 vN k2L2 (F r(Ω)) ≤ kvN k2L2 (Ω) ,
N
∀vN ∈ VN .
It holds that
aN (λN , λN ) = kwN k2L2 (Ω) ≥ c
1
1
kγ0 wN k2L2 (F r(Ω)) = c kγ0 λN k2L2 (F r(Ω)) .
N
N
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L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
Proposition 2. (Condition Number of the Matrix AN ) We have
cond(A) = O(N ),
where A is a square matrix associated to the problem (10).
Proof. Let A be a square inversible matrix N × N . We have
cond(A) = kAk .
A−1
,
where k.k is a matrix norm associated to canonical euclidean norm of RN . If
A is positive definite and symmetric then
cond(A) =
λmax
,
λmin
where λmax is the largest eigenvalue and λmin is the smallest eigenvalue of A.
By Theorem 7, there exist two constants α and β > 0 such that
α
1
kγ0 λN k2L2 (F r(Ω)) ≤ aN (λN , λN ) ≤ β kγ0 λN k2L2 (F r(Ω)) ,
N
Put
pmax =
sup
λN ∈MN −{0}
pmin =
aN (λN , λN )
kγ0 λN k2L2 (F r(Ω))
aN (λN , λN )
,
kγ
λN k2L2 (F r(Ω))
0
λN ∈MN −{0}
inf
then we have
pmax ≤ β and
pmin
≤
1
,
N
is the smallest eigenvalue of AN .
pmin ≥ α
where pmax is the largest eigenvale and pmin
Therefore, we have
1
∀λN ∈ MN .
pmax
N
⇒
≤ cN ⇒ cond(A) =
(N )
α
pmin
5.3. Algorithm (Conjugate Gradient Method Applied to Variational Problem (10)
We introduce the isomorphism rN : MN → Rp defined as follows :
µ N ∈ MN , µ N =
174
p
X
i=1
µi ϕ i ,
L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
rN = {µ1 , µ2 , ..., µP } ,
∀µN ∈ MN .
Then
aN (λN , µN ) = (AN rN λN , rN µN )N ,
and
Z
Z
Z
∇u0N ∇µN dx− w0N µN dx −
F r(Ω)
Ω
Ω
∀λN , µN ∈ MN
g2N µN dσ = (bN , rN µN )N , ∀µN ∈ MN ,
where (., .)N is the inner euclidien scalar of Rp and k.kN is the associated
norm.
Algorithm :
Step 1: k = 0, let λ0N ∈ MN be an arbitrary initial data.
0
gN
= AN rN λ0N − bN ,
0
d0N = gN
0
Step 2 : If kgN
kN ≤ ε stop, else put
ρn =
n
(dn
N , gN ) N
n
(AN dn
,
N dN ) N
= rN λnN − ρn dnN
rN λn+1
N
n+1
n
= gN
− ρn AN dnN
gN
and
βn =
n+1
n+1
, gN
)N
(gN
n , gn )
(gN
N N
n+1
+ βn dn+1
= gN
dn+1
N
N
Step 3 : k ← k + 1 and return to step 2.
Numerical Results :
Example 1. Consider the following problem :
∆2 u = −128π 4 cos(4πx) in Ω = ]−1, +1[2 ,
u = 0 on F r(Ω),
∂u
= 0 on F r(Ω).
∂η
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L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
Figure 1: exact solution u(x) = (sin(2πx))2
This problem has one and only one solution : u(x, y) = (sin(2πx))2 .
We have:
If we select the 3 rd choice of the space MN with
φ−1 = L1 − L0 ,
then we have
φN −1 = LN −1 ,
Figure 2: approach solution for N=12
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L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
Figure 3: Comparison between exact and approach solutions for N=12
Figure 4: approach solution for N=14
Figure 5: Comparison between exact and approach solutions for N=14
177
L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
Figure 6: approach solution for N=16
Figure 7: Comparison between exact and approach solutions for N=16
178
L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
Example 2. Consider the following problem :
∆2 u = 24(1 − x2 )2 + 24(1 − y 2 )2 + 32(3x2 − 1)(3y 2 − 1) in Ω = ]−1, +1[2 ,
u = 0 on F r(Ω),
∂u = 0 on F r(Ω).
∂η
This problem has one and only one solution: u(x, y) = (1 − x2 )2 (1 − y 2 )2 . We
have
Figure 8: exact solution
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L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
Figure 9: exact solution with contour
If we select the 1st choice of the space MN with
φ−1 = L0 ,
φN −1 = L1 ,
then we have
Figure 10: approach solution with line of contour for N=2
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L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
Figure 11: approach solution with lines of contour for N=4
6. Conclusion
We will say that developped methods are globally interesting when we
dispose beforehand a good and well program to resolve the Dirichlet approach
problem. One generalization of this problem is given in ([2]). An other more
intersting problem is to study the evolution problem and the estimate of error
for this last.
References
[1] L. Benaissa and N. Daili, Spectral Approximations of Stationary Dirichlet Problem for Harmonic Operator, Submited (2008), pp. 1-22.
[2] L. Benaissa and N. Daili, Spectral Approximations of Stationary Dirichlet Problem for Polyharmonic Operator and Decompositions, Preprint (2008),
pp. 1-50.
[3] C. Bernardi-Y. Maday, Approximation Spectrale de Probl`emes aux limites elliptiques, Springer-Verlag, Berlin, 199.
[4] R. Glowinski - O. Pironneau, Sur une m´ethode Quasi-Directe pour
l’Op´erateur Biharmonique et ses Applications `
a la R´esolution des Equations
de Navier-Stokes, Ann. Sc. Math. Qu´ebec, Vol.1, No. 2, (1977), pp. 231-245.
[5] J. W. McLaurin, A general coupled equation approach for solving, Siam.
Jour. Num. Anal.11 (1974), pp. 14-33.
[6] J. Shen, Efficient spectral-Galerkin method I. Direct solvers of second
and fourth-order using Legendre polynomials, Siam. Jour. Sci Comput. 5
181
L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
(1994), pp. 1489-1505.
Authors:
Lakhdar Benaissa
Hight school of commerce.
1 Rampe Salah Gharbi. Algiers. Algeria.
email: l [email protected]
Noureddine Daili
Department of Mathematics.
University of Ferhat Abbas.
19000 S´etif-Algeria
email: nourdaili [email protected]
182
No 17/2009
ABOUT BIHARMONIC PROBLEM VIA A SPECTRAL
APPROACH AND DECOMPOSITION TECHNIQUES
Lakhdar Benaissa, Noureddine Daili
Abstract. In this work, we describe a spectral method coupled with a
variational decomposition technique for solving a biharmonic equations. We
construct new spaces. Using one approximation by a spectral method we can
bring the resolution of Dirichlet problem for ∆2 to a finite number of Dirichlet
approach for -∆. Some new theoretic spectral approaches are given, numerical
solutions and illustrations are established to prove our theoretic study.
2000 Mathematics Subject Classification: 65N35, 65N30; 65N22.
Keywords and phrases: Biharmonic problem, spectral approach, approximation.
1. Introduction
Let Ω be a bounded domain of Rd (d = 1, 2, 3 in practice) of smooth boundary F r(Ω). We consider the following problem:
2
∆ u=f
in Ω,
u = g1
on F r(Ω) ;
(DP )
∂u
= g2 on F r(Ω) ,
∂η
where ∆2 = ∆(∆) is the biharmonic operator. The reduction of boundary
problems to equivalent problems is doing by several theoretic methods. We
are interested in this work to one direct method based on the Green formula.
This approach is used with success, by several authors, to resolve mathematical
modelling problems.
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L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
The solution of biharmonic problem (DP ) is studied by some authors and
by several methods, so by finite difference methods, finite element methods
and duality ...
J.W. McLaurin ([5], 1974) has given a decomposition technique for the
problem (DP ). He has proved that a solution of this problem by finite difference methods is equivalent to resolve a sequence of Dirichlet problems for the
operator ∆.
R. Glowinski and O. Pironneau ([4], 1977) have, only, established this decomposition method using the finite element methods correspondent to same
problem on a domain of R2 . They have proved that a solution of this problem
by this method is equivalent, also, to resolve a sequence of Dirichlet problems
for the operator ∆ and have given the error estimate:
ku − uh kH 1 (Ω) + k∆u + ϕh kL2 (Ω) =
(hk−1 ).
Our work consists to give a spectral approach to this problem and estimate
the error theoretically on new space. We illustrate our numerical approach by
numerical tests on some examples.
Indeed, if we resolve the biharmonic problem by spectral method based
on the direct variational formulation, then we can give as advantage of this
method the following :
i) we store one convergent approximation of the solution u in norm k.kH 1 (Ω)
in place of the norm k.kH 2 (Ω) ;
ii) We obtain a convergent approximation ϕN of −∆u.
However, we will think that, in spite of the simplicity of the space VN , the
practice problem is to compute (uN , ϕN ).
The basic idea in this case, consists to introduce a space MN ⊂ VN of
multipliers such that
VN = VN0 ⊕ MN
where
VN0 = {vN ∈ VN : vN = 0 on F r(Ω)}
In this paper, we prove that a resolution of Dirichlet problem for biharmonic
operator by Galerkin spectral method is equivalent to resolve a sequence of
Dirichlet problems for operator ∆ by the same method, and to resolve well
conditioned linear system. We prove some results of convergence which are
effecient and improve previously obtained results. We illustrate these results
by numerical trials.
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L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
2. A Continuous Problem
Let Ω be a bounded domain of Rd (d = 1, 2, 3 in practice) with smooth
boundary F r(Ω). Define the following problem:
2
∆ u=f
in Ω,
u = g1
on F r(Ω),
(DP )
∂u
= g2 on F r(Ω),
∂η
where g1 , g2 and f are given functions. This problem is equivalent to find
(u, w) such that
−∆w = f
in Ω,
−∆u = w on Ω,
(EP )
u = g1
on F r(Ω),
∂u
= g2 on F r(Ω).
∂η
3
1
We prove that if f ∈ L2 (Ω), g1 ∈ H 2 (F r(Ω)) and g2 ∈ H 2 (F r(Ω)) then (DP )
has one and only one solution in H 2 (Ω).
Theorem 1. [4] For all reals 0 ≤ ν ≤ µ we have
kukH µ (Ω) ≤ cN 2(µ−ν) kukH ν (Ω) ,
∀u ∈ VNd (Ω).
Theorem 2. [5] Define the space W 0 as follows:
W 0 = (v, ψ) ∈ H01 (Ω) × L2 (Ω), ∀µ ∈ H 1 (Ω), S((v, ψ), µ) = 0 ,
where
S((v, ψ), µ) =
Z
Ω
∇v∇µ dx −
Z
ψµ dx.
Ω
Then, a mapping : (v, ψ) ∈ W 0 → kψkL2 (Ω) is a norm on W 0 equivalent to the
norm:
(v, ψ) ∈ W 0 → (|v|2H 1 (Ω) + kψk2L2 (Ω) )1/2 .
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L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
Further
W 0 = (v, ψ) ∈ H02 (Ω) × L2 (Ω), − ∆v = ψ .
We have
1
Theorem 3. [6] Let λ ∈ H − 2 (F r(Ω)). The problem
2
∆ u = 0 in Ω,
−∆u = λ on F r(Ω),
u = 0 on F r(Ω),
(1)
has one and only one solution in H 2 (Ω)∩H01 (Ω), and the linear operator A
defined by
∂u
on F r(Ω)
Aλ = −
∂η
1
1
is an isomorphism from H − 2 (F r(Ω)) into H 2 (F r(Ω)). Also, the bilinear form
a(., .) defined by
a(λ, µ) =< Aλ, µ >
1
1
where < ., . > is the bilinear form of duality between H − 2 (F r(Ω)) and H 2 (F r(Ω)),
1
is continuous, symmetric and H − 2 (F r(Ω))-elliptic. Namely there exist an
α > 0 such that
a(µ, µ) ≥ α kµk2H − 21 (F r(Ω)) ,
1
∀µ ∈ H − 2 (F r(Ω)).
1
We will reduce the (DP ) problem to a variational equation in H − 2 (F r(Ω)).
Indeed, let w0 and u0 be solutions of
in Ω,
−∆w0 = f
(2)
w0 = 0
on F r(Ω);
−∆u0 = w0 in Ω,
(3)
u0 = g1 on F r(Ω).
We have then
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L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
Theorem 4. [4] Let u be a solution of (DP ). The trace λ of −∆u on
F r(Ω) is an unique solution of variational equation
< Aλ, µ >=<
∂u0
− g2 , µ > ,
∂η
1
1
∀µ ∈ H − 2 (F r(Ω)), λ ∈ H − 2 (F r(Ω)).
0
Remark 1. To determine ∂u
we resolve two Dirichlet problems (2), (3), and
∂η
we resolve more again two Dirichlet problems to obtain the couple (u, w).
3. Main Results
3.1. Approximation of (DP ) by Spectral Method
We consider the following finite dimension space:
VN = Span {L0 , L1 , ..., LN }
where Lk (x) are Legendre polynomials. Let
n
o
VN0 = vN ∈ VN : vN|F r(Ω) = 0
and let MN ⊂ VN be such that VN = VN0 ⊕ MN . We introduce the following
spaces :
and
(vN , ψN ) ∈ VN × VN , vN|F r(Ω) = g1 ,
WN =
;
R
R
R
∇vN ∇µN dx = Ω ψN µN dx + F r(Ω) g2 µN dσ, ∀µN ∈ VN
Ω
WN0
=
(vN , ψN ) ∈
VN0
× VN ,
Z
Ω
∇vN ∇µN dx −
Z
Ω
ψN µN dx = 0 , ∀µN ∈ VN
3.1.1. Choice of VN0 and Appropriate Basis for Galerkin Method
If we have a nonhomogenous Dirichlet problem associated to the operator
(∆) and posed on the space VN , we can define one transformation for which
this problem will be equivalent to one posed homogenous problem on VN0 .
The Galerkin approach consists plainly to replace the test functions space
by polynomials space.
The effectiveness of numerical method to resolve the linear system Au = F
wich has been given in the abstract form will be subordinate to :
161
.
L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
i) the way from which space VN0 (Ω) approaches the space V ;
ii) steepness and the simpleness calculus of coefficients aij and Fj ;
iii) steepness to resolve a linear system Au = F.
To satisfy the 1st criterion i), we will consider the space VN0 (Ω) of
enough large dimension.
To satisfy the 2nd and 3rd critera ii) and iii), it will require obtain
one sufficiently deep matrix A such that the linear system Au = F
does not enough at cost (in time and required space machine), and
such that there are not coefficients aij to compute.
What is to be done in the choice of a basis of VN0 (Ω) such that the
linear system to resolve will be possible ?
To answer this question, we need the following lemma :
Lemma 1. [6] Put
ck =
√ 1
,
4k+6
′
φk (x) = ck (Lk (x) − Lk+2 (x))
′
ajk = (φk (x), φj (x))
,
bjk = (φk (x), φj (x)), k, j = 0, 1, ..., N − 2.
Then,
ajk =
1 if k = j,
0 if k 6= j,
2
2
+ 2j+5
),
if k = j,
ck cj ( 2j+1
=
−ck cj ,
if k = j + 2,
0
else,
bkj = bjk
and
VN0 (I) = Span {φ0 (x), φ1 (x), ..., φN −2 (x)} ,
where (., .) is an L2 -inner product.
3.1.2. Choice of MN
We select MN ⊂ VN such that VN = VN0 ⊕ MN , where
VN0 = Span {φ0 , φ1 , ..., φN −2 } .
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L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
Let φ−1 , φN −1 be two functions such that φ−1 , φN −1 ∈ VN and φ−1 ,φ0 ,
φ1 , ..., φN −2 , φN −1 are linearly independent.
Proposition 1. For d = 1, the space MN is given by
MN = Span {φ−1 (x), φN −1 (x)} .
For d = 2, the space MN is given by
φ−1 (x)φj (y), φN −1 (x)φj (y),
MN = Span
φi (x)φ−1 (y), φi (x)φN −1 (y),
−1≤j ≤N −1
0≤i≤N −2
.
For d = 3, the space MN is given by
φi (x)φj (y)φ−1 (z), φi (x)φj (y)φN −1 (z), −1 ≤ i, j ≤ N − 1;
φi (x)φ−1 (y)φk (z), φi (x)φN −1 (y)φk (z), −1 ≤ i ≤ N − 1,
MN = Span
0 ≤ k ≤ N − 2;
φ−1 (x)φj (y)φk (z), φN −1 (x)φj (y)φk (z), 0 ≤ j, k ≤ N − 2
Remark 2. The choice of functions φ−1 , φN −1 is not unique, then the
choice of space MN is not unique.
Now, we give some choices of φ−1 , φN −1 :
Remark 3. i) φ−1 = L0 , φN −1 = L1
ii) φ−1 = L0 , φN −1 = LN −1 − L1
iii) φ−1 = L1 − L0 , φN −1 = LN −1 .
Numerical tests are proofs to the good choice.
We approache then (DP ) by
Find (uN , ϕN ) ∈ WN such that
(DP )N
jN (uN , ϕN ) ≤ jN (vN , ψN ), ∀(vN , ψN ) ∈ WN ,
where
1
jN (vN , ψN ) =
2
Z
2
Ω
|ψN | dx −
This problem has one and only one solution.
163
Z
Ω
f vN dx.
.
L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
4. Convergence Results
Consider the following problem:
2
∆ u=g
u=0
(DP )0
∂u
=0
∂η
in Ω,
on F r(Ω),
on F r(Ω),
which is equivalent to the following optimization problem:
Find u ∈ H02 (Ω) such that
(DP )1
J0 (u) ≤ J0 (v), ∀v ∈ H02 (Ω),
where
1
J0 (v) =
2
Z
2
|∆v| dx −
Ω
Z
gvdx.
Ω
Now, we prove some results of convergence.
Lemma 2. Let u be a solution of problem (DP )0 . Then there exist a
constant c > 0 independent of N such that
|u − uN |H 1 (Ω) + k∆u + ϕN kL2 (Ω) ≤ c(
inf
0
(vN , ψN )∈WN
(|u − vN |H 1 (Ω) + k∆u + ψN kL2 (Ω) )
+ inf k∆u + µN kH 1 (Ω) ).
µN ∈VN
Proof. Let u be a solution of problem (DP )0 , then
Z
Z
Z
gvdx, ∀v ∈ D(Ω)
∆u∆vdx =
− ∇v∇ (∆u) dx =
Ω
Ω
Ω
but D(Ω) is dense in H01 (Ω), then
Z
Z
gvdx, ∀v ∈ H01 (Ω).
− ∇v∇(∆u)dx =
Ω
Ω
We have
S((v, ψ), µ) =
Z
Ω
∇v∇µdx −
164
Z
Ω
ψµdx,
L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
and
S((v, ψ), −∆u) = −
Z
Ω
∇v∇(∆u)dx +
Z
ψ∆udx =
Z
gvdx +
ψ∆udx.
Ω
Ω
Ω
Z
Let (vN , ψN ) ∈ WN0 and µN ∈ VN , then
R
S((uN − vN , ϕN − ψN ), ∆u + µN ) = Ω ∇(uN − vN )∇(∆u + µN )dx
−
R
Ω
(ϕN − ψN )(∆u + µN )dx
= S((uN − vN , ϕN − ψN ), ∆u) + S((uN − vN , ϕN − ψN ), µN )
R
=−
g(uN − vN )dx −
Ω
+
=−
R
Ω
+
Or
Z
Ω
R
Ω
∇uN ∇µN dx −
∇uN ∇µN dx =
Z
Ω
(ϕN − ψN )∆udx
∇(uN − vN )∇µN dx −
Ω
(ϕN − ψN )∆udx −
R
R
R
Ω
R
Ω
guN dx +
R
Ω
∇vN ∇µN dx −
ϕN µN dx and
Z
Ω
(ϕN − ψN )µN dx
gvN dx
R
Ω
ϕN µN dx +
ϕN ψN dx =
Ω
Ω
R
Z
R
Ω
ψN µN dx.
gvN dx,
Ω
then
Z
Ω
(ϕN − ψN )(∆u + ϕN )dx = −S((uN − vN , ϕN − ψN ), ∆u + µN )
By continuity of S ((., .) , .) , we have
R
(ϕN − ψN )(∆u + ϕN )dx ≤ c(|uN − vN | 1 k∆u + µN k 1
H (Ω)
H (Ω)
Ω
+ kϕN − ψN kL2 (Ω) k∆u + µN kH 1 (Ω) )
Using Theorem 2, it holds,
|v|H 1 (Ω) ≤ c(Ω) kψkL2 (Ω)
165
L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
and
R
(ϕN − ψN )(∆u + ϕN )dx ≤ c(c(Ω) kϕN − ψN k 2 k∆u + µN k 1
L (Ω)
H (Ω)
Ω
+ kϕN − ψN kL2 (Ω) k∆u + µN kH 1 (Ω) )
≤ d(Ω) kϕN − ψN kL2 (Ω) k∆u + µN kH 1 (Ω) ,
where d(Ω) = max(cc(Ω), c). Thus
R
R
kϕN − ψN k2L2 (Ω) = Ω (ϕN − ψN )(∆u + ϕN )dx − Ω (ϕN − ψN )(∆u + ψN )dx
kϕN − ψN k2L2 (Ω) ≤ d(Ω) kϕN − ψN kL2 (Ω) k∆u + µN kH 1 (Ω)
+ kϕN − ψN kL2 (Ω) k∆u + ψN kL2 (Ω) .
It holds that
kϕN − ψN kL2 (Ω) ≤ d(Ω) k∆u + µN kH 1 (Ω) + k∆u + ψN kL2 (Ω)
(4)
and
|u − uN |H 1 (Ω) + k∆u + ϕN kL2 (Ω) ≤ |u − vN |H 1 (Ω) + |vN − uN |H 1 (Ω)
+ k∆u + ψN kL2 (Ω) + kϕN − ψN kL2 (Ω)
≤ |u − vN |H 1 (Ω) + k∆u + ψN kL2 (Ω) + (1 + c(Ω)) kϕN − ψN kL2 (Ω) .
Using inequality (4), we obtain
|u − uN |H 1 (Ω) + k∆u + ϕN kL2 (Ω) ≤ (|u − vN |H 1 (Ω) + k∆u + ψN kL2 (Ω)
+(1 + c(Ω))) · (d(Ω) k∆u + µN kH 1 (Ω) + k∆u + ψN kL2 (Ω) ),
hence
|u − uN |H 1 (Ω) + k∆u + ϕN kL2 (Ω) ≤ |u − vN |H 1 (Ω) + (2 + c(Ω)) k∆u + ψN kL2 (Ω)
+(1 + c(Ω))d(Ω) k∆u + µN kH 1 (Ω)
≤ c∗ (|u − vN |H 1 (Ω) + k∆u + ψN kL2 (Ω) + k∆u + µN kH 1 (Ω) ),
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L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
where
c∗ = max(2 + c(Ω), (1 + c(Ω))d(Ω)).
Therefore, it holds
|u − uN |H 1 (Ω) + k∆u + ϕN kL2 (Ω) ≤ c∗ (
inf
0
(vN ,ψN )∈WN
(|u − vN |H 1 (Ω) + k∆u + ψN kL2 (Ω) )
+ inf k∆u + µN kH 1 (Ω) ).
µN ∈VN
Lemma 3. Let u ∈ H 2 (Ω) be a solution of optimization problem (DP )0 .
If u ∈ H k+2 (Ω), k ≥ 2, then there exist a constant c > 0 independent of N
such that
|u − uN |H 1 (Ω) + k∆u + ϕN kL2 (Ω) ≤ cN 1−k (|u|H k+2 (Ω) + |∆u|H k (Ω) )
Proof. Let (vN , ψN ) ∈ WN0 and µN ∈ VN . Put
νN = µN + ψN , νN ∈ VN .
Then, we have
Z
Ω
Or
S((vN , ψN ), νN ) = 0,
Z
Z
∆uνN dx = − ∇u∇νN dx +
F r(Ω)
Ω
(5)
∂u
νN dσ
∂η
∂u
|F r(Ω) = 0, so one has
∂η
Z
Z
∆uνN dx = − ∇u∇νN dx.
Ω
Ω
By (5) and (6), we have
Z
Z
(∆u + ψN )νN dx =
∇(vN − u)∇νN dx
Ω
Ω
Z
(∆u + ψN )νN dx ≤ |u − vN | 1 |νN | 1 .
H (Ω)
H (Ω)
Ω
By Theorem 1, we obtain
|νN |H 1 (Ω) ≤ cN 2 kνN kL2 (Ω)
167
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L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
so
Z
(∆u + ψN )νN dx ≤ cN 2 |u − vN | 1 kνN k 2
H (Ω)
L (Ω)
Ω
And
kνN k2L2 (Ω) =
R
Ω
(µN − ∆u)νN dx +
R
Ω
(∆u + ψN )νN dx
≤ k∆u − µN kL2 (Ω) kνN kL2 (Ω) + cN 2 |u − vN |H 1 (Ω) kνN kL2 (Ω)
Hence
⇒ kνN kL2 (Ω) ≤ k∆u − µN kL2 (Ω) + cN 2 |u − vN |H 1 (Ω).
k∆u + ψN kL2 (Ω) = k∆u + νN − µN kL2 (Ω)
≤ k∆u − µN kL2 (Ω) + kνN kL2 (Ω) ≤ 2 k∆u − µN kL2 (Ω) + cN 2 |u − vN |H 1 (Ω) .
Then
|u − vN |H 1 (Ω) + k∆u + ψN kL2 (Ω) ≤ (1 + cN 2 ) |u − vN |H 1 (Ω) + 2 k∆u − µN kL2 (Ω) ,
∀ (vN , ψN ) ∈ WN0 , ∀µN ∈ VN .
Thus
inf
0
(vN ,ψN )∈WN
(|u − vN |H 1 (Ω) + k∆u + ψN kL2 (Ω) ) ≤ (1 + cN 2 ) inf 0 |u − vN |H 1 (Ω)
vN ∈VN
+2 inf k∆u − µN kL2 (Ω) .
µN ∈VN
Lemma 2 implies
|u − uN |H 1 (Ω) + k∆u + ϕN kL2 (Ω) ≤ c((1 + cN 2 ) inf 0 |u − vN |H 1 (Ω)
vN ∈VN
+ inf k∆u − µN kH 1 (Ω) ).
µN ∈VN
Further, we have
inf |u − vN |H 1 (Ω) ≤ cN −1−k |u|H k+2 (Ω)
vN ∈VN0
inf k∆u + µN kH 1 (Ω) ≤ cN 1−k |∆u|H k (Ω) ,
µN ∈VN
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L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
then, it holds
|u − uN |H 1 (Ω) + k∆u + ϕN kL2 (Ω) ≤ c((1 + cN 2 )cN −1−k |u|H k+2 (Ω) + cN 1−k |∆u|H k (Ω) )
≤ cN 1−k (|u|H k+2 (Ω) + |∆u|H k (Ω) ).
Theorem 5. Le u ∈ H 2 (Ω) be a solution of a problem (DP ), then for all
integer k ≥ 2, if u ∈ H k+2 (Ω) we have
ku − uN kH 1 (Ω) + k∆u + ϕN kL2 (Ω) ≤ cN 1−k kukH k+2 (Ω)
If u ∈ H k+1 (Ω), k ≥ 3, then
ku − uN kL2 (Ω) +
1
k∆u + ϕN kL2 (Ω) ≤ cN −1−k kukH k+1 (Ω)
N2
Proof. We will reduce the problem (P D)N to one variational problem in
MN . Indeed, let aN (., .) : MN × MN → R be a bilinear form defined for
λN ∈ MN by
Z
∇wN ∇vN dx = 0 , ∀vN ∈ VN0 , wN − λN ∈ VN0 ;
(7)
Ω
Z
Ω
∇uN ∇vN dx =
aN (λN , µN ) =
Z
Ω
Z
Ω
wN vN dx , ∀vN ∈ VN0 , uN ∈ VN0 ;
wN µN dx −
Z
Ω
∇uN ∇µN dx,
∀µN ∈ MN .
(8)
(9)
Lemma 4. A bilinear form aN (., .) is positive definite and symmetric on
MN × MN . Moreover
Z
w1N w2N dx, ∀ λ1N , λ2N ∈ MN ,
aN (λ1N , λ2N ) =
Ω
where w1N , w2N are solutions of (7) associated to λ1N , λ2N .
Proof. By definition one has
Z
Z
w1N λ2N dx − ∇u1N ∇λ2N dx, ∀µN ∈ MN ,
aN (λ1N , λ2N ) =
Ω
Ω
where w1N , w2N are solutions of (7) and (8) respectively.
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L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
Put λ2N = (λ2N − w2N ) + w2N , then
R
R
R
aN (λ1N , λ2N ) = Ω w1N w2N dx − Ω ∇u1N ∇w2N dx + Ω ∇u1N ∇(w2N − λ2N ) dx
−
R
Ω
w1N ∇(w2N − λ2N ) dx
One has u1N ∈ VN0 , using (7) one has
Z
Z
w1N ∇(w2N − λ2N ) dx.
∇u1N ∇(w2N − λ2N ) dx =
Ω
Ω
Then
aN (λ1N , λ2N ) =
Z
w1N w2N dx,
Ω
∀ λ1N , λ2N ∈ MN .
It is obvious that aN (., .) is symmetric and coercive.
Theorem 6. Let (uN , wN ) be a solution of (DP )N and λN the component of wN in MN . Then λN is an unique solution of the linear variational
problem:
R
R
R
aN (λN , µN ) = Ω ∇u0N ∇µN dx − Ω w0N µN dx − F r(Ω) g2N µN dσ,
(10)
∀µN ∈ MN , λN ∈ MN
where w0N is a solution of
Z
Z
f vN dx , ∀vN ∈ VN0 , w0N ∈ VN0
∇w0N ∇vN dx =
and u0N is a solution of
R
R
Ω ∇u0N ∇vN dx = Ω w0N vN dx , ∀vN ∈ VN0 ,
(11)
Ω
Ω
u0N = g1
on
F r(Ω).
Proof. We have
R
R
uN ∇µN dx
aN (λN , µN ) = Ω w¯N µN dx − Ω ∇¯
=
R
(wN − w0N )µN dx −
Ω
=
R
∇u0N ∇µN dx −
Ω
R
R
Ω
∇(uN − u0N )∇µN dx
R
R
w µ dx).
∇u
∇µ
dx
−
w
µ
dx
−
(
N
N
0N
N
Ω N N
Ω
Ω
170
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L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
But (uN , wN ) ∈ WN , hence
Z
Z
Z
∇uN ∇µN dx − wN µN dx =
Ω
Ω
g2N µN dσ,
F r(Ω)
∀µN ∈ MN
Therefore
aN (λN , µN ) =
Z
Ω
Z
Z
∇u0N ∇µN dx − w0N µN dx−
Ω
g2N µN dσ,
F r(Ω)
∀µN ∈ MN .
Existence and uniqueness of a solution of (10) is a direct consequence from LaxMilgram Lemma. A system is symmetric because a form aN (., .) is symmetric.
Remak 4. The problem (10) is equivalent to a system from which associated matrix is positive definite and symmetric.
Now we prove a problem (10) is equivalent to a system from which associated matrix is positive definite.
5. Numerical Approach
5.1. Construction of Linear System Associated to the Variational
Problem (10)
Put
lN (µN ) =
Z
Ω
∇u0N ∇µN dx −
Z
Ω
w0N µN dx −
Z
g2 µN dσ,
F r(Ω)
∀µN ∈ MN
Then solve a problem (10) is equivalent to resolve the following system :
aN (λN , µN ) = lN (µN ),
∀µN ∈ MN , λN ∈ MN .
Define the set
{−1, N − 1}
(−1,
j),
(N
−
1,
j)
,
−1
≤
j
≤
N
−
1
,
(i, −1), (i, N − 1) ,
0≤i≤N −2
Id =
(i, j, −1), (i, j, N − 1) , −1 ≤ i, j ≤ N − 1,
(i, −1, k), (i, N − 1, k) , −1 ≤ i ≤ N − 1, 0 ≤ k ≤ N − 2
(−1, j, k), (N − 1, j, k), 0 ≤ j, k ≤ N − 2
171
(13)
if d = 1
if d = 2
, if d = 3,
L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
and
φid = φi1 φi2 · ... · φid
d = 1, 2, 3
where id = (i1 , i2 , ..., id ). So (13) is equivalent to
d
aN (λN , φid ) = lN (φid ) , i ∈ Id
P
λj d φj d .
λN =
(14)
j d ∈Id
Namely
X
λj d aN (φj d , φkd ) = lN (φkd ),
j d ∈Id
and
2,
4N,
p = Card(Id ) = dim(MN ) =
6N 2 + 2,
k d ∈ Id
(15)
if d = 1,
if d = 2,
if d = 3.
5.2. Computation of the Matrix AN
Denote BN the basis of MN and wj d N , uj d N solutions respectively of
R
wj d N ∈ VN , wj d N − φj d ∈ VN0 ;
∇wj d N ∇vN dx = 0, ∀vN ∈ VN0 ,
Ω
R
∇uj d N ∇vN dx =
Ω
further we have
R
Ω
aid j d = aN (φj d , φid ) =
wj d N vN dx, ∀vN ∈ VN0 ,
Z
Ω
wj d N φid dx −
Z
Ω
uj d N ∈ VN0 ,
∇uj d N ∇φid dx,
id , j d ∈ Id .
It results that a computation of a matrix AN requires the computation of 2p
Dirichlet problems for an operator ∆.
We have
Theorem 7. For all integer N ≥ 1, we have
α
1
kγ0 λN k2L2 (F r(Ω)) ≤ aN (λN , λN ) ≤ β kγ0 λN k2L2 (F r(Ω)) ,
N
where α and β are independents of N and λN .
172
∀λN ∈ MN ,
L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
Proof. Let λN ∈ MN , we have
aN (λN , λN ) =
Z
Ω
2
wN
dx,
where wN is a solution of (7). Let w˜N ∈ H 1 (Ω) be a solution of
∆w˜N = 0 in Ω,
w˜N = λN on F r (Ω) ,
then one has
p
aN (λN , λN ) = kwN kL2 (Ω)
≤ kwN − w˜N kL2 (Ω) + kw˜N kL2 (Ω) = kwN − w˜N kL2 (Ω) +
√
< Aγ0 λN , γ0 λN >.
Put |A| = kAkLc(L2 (F r(Ω)),L2 (F r(Ω))) , it holds that
|< Aλ, µ >| ≤ |A| kλkL2 (F r(Ω)) kµkL2 (F r(Ω)) ,
⇒
∀ λ, µ ∈ L2 (F r(Ω))
p
p
aN (λN , λN ) ≤ kwN − w˜N kL2 (Ω) + |A| kγ0 λN kL2 (F r(Ω)) .
and we have
3
kwN − w˜N kL2 (Ω) ≤ cN − 2 kw˜N kH 23 (Ω) ,
and
kw˜N kH 32 (Ω) ≤ c kλN kH 1 (F r(Ω))
kλN kH 1 (F r(Ω)) ≤ cN kλN kL2 (F r(Ω)) .
Namely
p
And
c
aN (λN , λN ) ≤ c kγ0 λN kL2 (F r(Ω)) .
1
kγ0 vN k2L2 (F r(Ω)) ≤ kvN k2L2 (Ω) ,
N
∀vN ∈ VN .
It holds that
aN (λN , λN ) = kwN k2L2 (Ω) ≥ c
1
1
kγ0 wN k2L2 (F r(Ω)) = c kγ0 λN k2L2 (F r(Ω)) .
N
N
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L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
Proposition 2. (Condition Number of the Matrix AN ) We have
cond(A) = O(N ),
where A is a square matrix associated to the problem (10).
Proof. Let A be a square inversible matrix N × N . We have
cond(A) = kAk .
A−1
,
where k.k is a matrix norm associated to canonical euclidean norm of RN . If
A is positive definite and symmetric then
cond(A) =
λmax
,
λmin
where λmax is the largest eigenvalue and λmin is the smallest eigenvalue of A.
By Theorem 7, there exist two constants α and β > 0 such that
α
1
kγ0 λN k2L2 (F r(Ω)) ≤ aN (λN , λN ) ≤ β kγ0 λN k2L2 (F r(Ω)) ,
N
Put
pmax =
sup
λN ∈MN −{0}
pmin =
aN (λN , λN )
kγ0 λN k2L2 (F r(Ω))
aN (λN , λN )
,
kγ
λN k2L2 (F r(Ω))
0
λN ∈MN −{0}
inf
then we have
pmax ≤ β and
pmin
≤
1
,
N
is the smallest eigenvalue of AN .
pmin ≥ α
where pmax is the largest eigenvale and pmin
Therefore, we have
1
∀λN ∈ MN .
pmax
N
⇒
≤ cN ⇒ cond(A) =
(N )
α
pmin
5.3. Algorithm (Conjugate Gradient Method Applied to Variational Problem (10)
We introduce the isomorphism rN : MN → Rp defined as follows :
µ N ∈ MN , µ N =
174
p
X
i=1
µi ϕ i ,
L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
rN = {µ1 , µ2 , ..., µP } ,
∀µN ∈ MN .
Then
aN (λN , µN ) = (AN rN λN , rN µN )N ,
and
Z
Z
Z
∇u0N ∇µN dx− w0N µN dx −
F r(Ω)
Ω
Ω
∀λN , µN ∈ MN
g2N µN dσ = (bN , rN µN )N , ∀µN ∈ MN ,
where (., .)N is the inner euclidien scalar of Rp and k.kN is the associated
norm.
Algorithm :
Step 1: k = 0, let λ0N ∈ MN be an arbitrary initial data.
0
gN
= AN rN λ0N − bN ,
0
d0N = gN
0
Step 2 : If kgN
kN ≤ ε stop, else put
ρn =
n
(dn
N , gN ) N
n
(AN dn
,
N dN ) N
= rN λnN − ρn dnN
rN λn+1
N
n+1
n
= gN
− ρn AN dnN
gN
and
βn =
n+1
n+1
, gN
)N
(gN
n , gn )
(gN
N N
n+1
+ βn dn+1
= gN
dn+1
N
N
Step 3 : k ← k + 1 and return to step 2.
Numerical Results :
Example 1. Consider the following problem :
∆2 u = −128π 4 cos(4πx) in Ω = ]−1, +1[2 ,
u = 0 on F r(Ω),
∂u
= 0 on F r(Ω).
∂η
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L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
Figure 1: exact solution u(x) = (sin(2πx))2
This problem has one and only one solution : u(x, y) = (sin(2πx))2 .
We have:
If we select the 3 rd choice of the space MN with
φ−1 = L1 − L0 ,
then we have
φN −1 = LN −1 ,
Figure 2: approach solution for N=12
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L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
Figure 3: Comparison between exact and approach solutions for N=12
Figure 4: approach solution for N=14
Figure 5: Comparison between exact and approach solutions for N=14
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L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
Figure 6: approach solution for N=16
Figure 7: Comparison between exact and approach solutions for N=16
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L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
Example 2. Consider the following problem :
∆2 u = 24(1 − x2 )2 + 24(1 − y 2 )2 + 32(3x2 − 1)(3y 2 − 1) in Ω = ]−1, +1[2 ,
u = 0 on F r(Ω),
∂u = 0 on F r(Ω).
∂η
This problem has one and only one solution: u(x, y) = (1 − x2 )2 (1 − y 2 )2 . We
have
Figure 8: exact solution
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L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
Figure 9: exact solution with contour
If we select the 1st choice of the space MN with
φ−1 = L0 ,
φN −1 = L1 ,
then we have
Figure 10: approach solution with line of contour for N=2
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L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
Figure 11: approach solution with lines of contour for N=4
6. Conclusion
We will say that developped methods are globally interesting when we
dispose beforehand a good and well program to resolve the Dirichlet approach
problem. One generalization of this problem is given in ([2]). An other more
intersting problem is to study the evolution problem and the estimate of error
for this last.
References
[1] L. Benaissa and N. Daili, Spectral Approximations of Stationary Dirichlet Problem for Harmonic Operator, Submited (2008), pp. 1-22.
[2] L. Benaissa and N. Daili, Spectral Approximations of Stationary Dirichlet Problem for Polyharmonic Operator and Decompositions, Preprint (2008),
pp. 1-50.
[3] C. Bernardi-Y. Maday, Approximation Spectrale de Probl`emes aux limites elliptiques, Springer-Verlag, Berlin, 199.
[4] R. Glowinski - O. Pironneau, Sur une m´ethode Quasi-Directe pour
l’Op´erateur Biharmonique et ses Applications `
a la R´esolution des Equations
de Navier-Stokes, Ann. Sc. Math. Qu´ebec, Vol.1, No. 2, (1977), pp. 231-245.
[5] J. W. McLaurin, A general coupled equation approach for solving, Siam.
Jour. Num. Anal.11 (1974), pp. 14-33.
[6] J. Shen, Efficient spectral-Galerkin method I. Direct solvers of second
and fourth-order using Legendre polynomials, Siam. Jour. Sci Comput. 5
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L. Benaissa, N. Daili - About Biharmonic Problem via a Spectral Approach
(1994), pp. 1489-1505.
Authors:
Lakhdar Benaissa
Hight school of commerce.
1 Rampe Salah Gharbi. Algiers. Algeria.
email: l [email protected]
Noureddine Daili
Department of Mathematics.
University of Ferhat Abbas.
19000 S´etif-Algeria
email: nourdaili [email protected]
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